# Consecutive Dice Throws

## Problem

What is the expected number of times one must roll a fair dice before getting $$2$$ consecutive $$6$$’s?

## Solution

According to the geometric distribution, we can expect $$6$$ rolls before landing the first $$6$$. At that point, there is a $$\frac{1}{6}$$ chance that the next roll is a $$6$$, which would result in two consecutive $$6$$’s. On the other hand, there is a $$\frac{5}{6}$$ chance of landing a number that is not $$6$$, at which point we must reset our count. This formulation can be represented by the following recurrence, where $$X$$ is a random variable representing the number of rolls for $$2$$ consecutive $$6$$’s:

$E[X] = 6 + \frac{1}{6}(1) + \frac{5}{6}(1+E[X])$ $E[X] = 42$

Thus, the expected number of rolls to land $$2$$ consecutive $$6$$’s is $$42$$.

This approach can be generalized to an arbitrary number of consecutive $$6$$’s. For example, suppose $$Y$$ represents the number of rolls for $$3$$ consecutive $$6$$’s. Knowing that $$E[X] = 42$$, we can write a recurrence for $$E[Y]$$ as follows:

$E[Y] = E[X] + \frac{1}{6}(1) + \frac{5}{6}(1+E[Y])$ $E[Y] = 258$